3倍角の公式

複素数を使う：ド・モアブルの定理

ド・モアブルの定理 - Wikipediaを使う．

$e^{i3\theta} = (e^{i\theta})^{3}$だから

\begin{aligned}
e^{i3\theta} & = \cos 3\theta + i \sin 3\theta \\
& \!\!\!\!\!\!\!\! \| \\
(e^{i\theta})^{3}
&=(\cos\theta + i \sin\theta)^{3} \\
&=\cos^{3}\theta
+ 3\cos^{2}\theta \cdot i \sin\theta \\
&\quad
+ 3\cos\theta \cdot (i \sin\theta)^{2}
+ (i \sin\theta)^{3} \\
&= [\cos^{3}\theta - 3\cos\theta \sin^{2}\theta] \\
&\quad
+ i [3\cos^{2}\theta \sin\theta - \sin^{3}\theta]
\end{aligned}

となる（$(x + y)^{3}$を展開したときのの係数（二項係数）はパスカルの三角形 - Wikipediaで1, 3, 3, 1とわかる）．

実部を比較すれば

\begin{aligned}
\cos 3\theta
&=\cos^{3}\theta - 3\cos\theta \overbrace{\sin^{2}\theta}^{\mathrlap{=1 - \cos^{2}\theta}} \\
&=4\cos^{3}\theta - 3\cos\theta
\end{aligned}

虚部を比較すれば

\begin{aligned}
\sin 3\theta
&=3 \overbrace{\cos^{2}\theta}^{\mathrlap{=1 - \sin^{2}\theta}} \sin\theta - \sin^{3}\theta \\
&=3\sin\theta - 4 \sin^{3}\theta
\end{aligned}

加法定理を使う

\begin{aligned}
\cos 3\theta
&= \cos (2\theta + \theta) \\
&= \cos 2\theta \cos\theta - \sin 2\theta \sin\theta \\
&= (\cos^{2}\theta - \sin^{2}\theta) \cos\theta
- 2\sin\theta \cos\theta \sin\theta \\
&=\cos^{3}\theta -3\overbrace{\sin^{2}\theta}^{\mathrlap{=1 - \cos^{2}\theta}} \cos\theta \\
&=4\cos^{3}\theta - 3\cos\theta
\end{aligned}

\begin{aligned}
\sin 3\theta
&= \sin (2\theta + \theta) \\
&= \sin 2\theta \cos\theta + \cos 2\theta \sin\theta \\
&= 2\sin\theta \cos^{2}\theta + (\cos^{2}\theta - \sin^{2}\theta)\sin\theta \\
&= 3\sin\theta \overbrace{\cos^{2}\theta}^{\mathrlap{=1 - \sin^{2}\theta}} - \sin^{3}\theta \\
&=3\sin\theta - 4 \sin^{3}\theta
\end{aligned}

回転行列を使う

回転行列（回転行列の表式と導出（2次元・3次元） - Notes_JP）を使う．
（※ 数学ガール／ガロア理論の方法）

\begin{aligned}
&
\begin{pmatrix}
\cos 3\theta & -\sin 3\theta \\
\sin 3\theta & \cos 3\theta
\end{pmatrix} \\
&=
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}^{3} \\
&=
\begin{pmatrix}
\cos^{2}\theta -\sin^{2}\theta & -2\sin\theta \cos\theta \\
2\sin\theta \cos\theta & \cos^{2}\theta -\sin^{2}\theta
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix} \\
&=
\begin{pmatrix}
\cos^{3}\theta - 3\sin^{2}\theta \cos\theta & -3\sin\theta \cos^{2}\theta + \sin^{3}\theta\\
3\sin\theta \cos^{2}\theta -\sin^{3}\theta & \cos^{3}\theta - 3\sin^{2}\theta \cos\theta
\end{pmatrix}
\end{aligned}

成分を比較すれば，

\begin{aligned}
\cos 3\theta
&=\cos^{3}\theta - 3\cos\theta \overbrace{\sin^{2}\theta}^{\mathrlap{=1 - \cos^{2}\theta}} \\
&=4\cos^{3}\theta - 3\cos\theta
\end{aligned}

\begin{aligned}
\sin 3\theta
&=3 \overbrace{\cos^{2}\theta}^{\mathrlap{=1 - \sin^{2}\theta}} \sin\theta - \sin^{3}\theta \\
&=3\sin\theta - 4 \sin^{3}\theta
\end{aligned}